Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{y}{4y^2 + 18y} \times \dfrac{2y + 9}{6} $
When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ y \times (2y + 9) } { (4y^2 + 18y) \times 6 } $ $ r = \dfrac {y (2y + 9)} {6 \times 2y(2y + 9)} $ $ r = \dfrac{y(2y + 9)}{12y(2y + 9)} $ We can cancel the $2y + 9$ so long as $2y + 9 \neq 0$ Therefore $y \neq -\dfrac{9}{2}$ $r = \dfrac{y \cancel{(2y + 9})}{12y \cancel{(2y + 9)}} = \dfrac{y}{12y} = \dfrac{1}{12} $